a) Divide $$ x+2 x^{2}+3 x+4 $$ by $$ x+2 $$ b) Divide $$ 2 x^{3}-x^{2}+3 x-2 $$ by $$ x+3 $$ c) Divide $$ a^{3}+64 b^{3} $$ by $$ a+4 b $$ 2. Find the remainder when $$ x^{3}+3 x^{2}-x+7 $$ is divided by $$ 3 x-1 $$.

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**(a) Divide $$ x + 2x^{2} + 3x + 4 $$ by $$ x + 2 $$**

First, rewrite the dividend in standard form (from highest degree to constant term):

$$
2x^{2} + (x + 3x) + 4 = 2x^{2} + 4x + 4.
$$

Now, perform the division:

1. Divide the leading term of the dividend by the leading term of the divisor:
   $$
   \frac{2x^{2}}{x} = 2x.
   $$
   
2. Multiply the divisor by $$ 2x $$:
   $$
   2x(x + 2) = 2x^{2} + 4x.
   $$

3. Subtract this from the dividend:
   $$
   (2x^{2} + 4x + 4) - (2x^{2} + 4x) = 4.
   $$

Since $$ 4 $$ is a constant and cannot be divided further by $$ x+2 $$, it is the remainder.

**Result:**  
Quotient: $$ 2x $$  
Remainder: $$ 4 $$

---

**(b) Divide $$ 2x^{3} - x^{2} + 3x - 2 $$ by $$ x + 3 $$**

Perform long division step by step:

1. Divide the leading term:
   $$
   \frac{2x^{3}}{x} = 2x^{2}.
   $$
   
2. Multiply the divisor by $$ 2x^{2} $$:
   $$
   2x^{2}(x + 3) = 2x^{3} + 6x^{2}.
   $$
   
3. Subtract:
   $$
   (2x^{3} - x^{2}) - (2x^{3} + 6x^{2}) = -7x^{2}.
   $$
   Bring down the next term:
   $$
   -7x^{2} + 3x.
   $$

4. Divide:
   $$
   \frac{-7x^{2}}{x} = -7x.
   $$
   
5. Multiply the divisor by $$ -7x $$:
   $$
   -7x(x + 3) = -7x^{2} - 21x.
   $$
   
6. Subtract:
   $$
   (-7x^{2} + 3x) - (-7x^{2} - 21x) = 3x + 21x = 24x.
   $$
   Bring down the final term:
   $$
   24x - 2.
   $$

7. Divide:
   $$
   \frac{24x}{x} = 24.
   $$
   
8. Multiply the divisor by $$ 24 $$:
   $$
   24(x + 3) = 24x + 72.
   $$
   
9. Subtract:
   $$
   (24x - 2) - (24x + 72) = -74.
   $$

**Result:**  
Quotient: $$ 2x^{2} - 7x + 24 $$  
Remainder: $$-74$$

---

**(c) Divide $$ a^{3} + 64b^{3} $$ by $$ a + 4b $$**

Recognize that $$ 64b^{3} = (4b)^{3} $$ so the expression is a sum of cubes. The sum of cubes formula is:

$$
u^3 + v^3 = (u + v)(u^2 - uv + v^2)
$$

Here, take $$ u = a $$ and $$ v = 4b $$. Then,

$$
a^{3} + (4b)^{3} = (a + 4b)\left(a^2 - a(4b) + (4b)^2\right)
$$

Simplify:

$$
a^2 - 4ab + 16b^2.
$$

Since $$ a + 4b $$ is a factor, the division yields exactly the other factor.

**Result:**  
Quotient: $$ a^2 - 4ab + 16b^2 $$  
Remainder: $$ 0 $$

---

**(2) Find the remainder when $$ x^{3} + 3x^{2} - x + 7 $$ is divided by $$ 3x - 1 $$**

To use the Remainder Theorem for a divisor that is not monic, first solve $$ 3x - 1 = 0 $$:

$$
3x - 1 = 0 \quad \Longrightarrow \quad x = \frac{1}{3}.
$$

Evaluate the polynomial at $$ x = \frac{1}{3} $$:

$$
\begin{aligned}
f\left(\frac{1}{3}\right) &= \left(\frac{1}{3}\right)^3 + 3\left(\frac{1}{3}\right)^2 - \frac{1}{3} + 7 \
&= \frac{1}{27} + 3\left(\frac{1}{9}\right) - \frac{1}{3} + 7 \
&= \frac{1}{27} + \frac{1}{3} - \frac{1}{3} + 7 \
&= \frac{1}{27} + 7 \
&= \frac{1 + 189}{27} \
&= \frac{190}{27}.
\end{aligned}
$$

**Result:**  
Remainder: $$ \frac{190}{27} $$
**(a) Divide $$ x + 2x^{2} + 3x + 4 $$ by $$ x + 2 $$:** Quotient: $$ 2x $$ Remainder: $$ 4 $$ **(b) Divide $$ 2x^{3} - x^{2} + 3x - 2 $$ by $$ x + 3 $$:** Quotient: $$ 2x^{2} - 7x + 24 $$ Remainder: $$-74$$ **(c) Divide $$ a^{3} + 64b^{3} $$ by $$ a + 4b $$:** Quotient: $$ a^{2} - 4ab + 16b^{2} $$ Remainder: $$ 0 $$ **(2) Remainder when $$ x^{3} + 3x^{2} - x + 7 $$ is divided by $$ 3x - 1 $$:** Remainder: $$ \frac{190}{27} $$

a) Divide \( x+2 x^{2}+3 x+4 \) by \( x+2 \) b) Divide \( 2 x^{3}-x^{2}+3 x-2 \) by \( x+3 \) c) Divide \( a^{3}+64 b^{3} \) by \( a+4 b \) 2. Find the remainder when \( x^{3}+3 x^{2}-x+7 \) is divided by \( 3 x-1 \).

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